Math help Please?

Question

[sqrt(-2)+18] = 4


Also :


[sqrtx-1] = x-3



HELP !!!

Answer 1

The first question doesn't make sense, because there's no x variable.

The second, however, presuming it's this:

sqrt(x - 1) = x - 3

Your first step would be to square both sides. This eliminates the square root.

x - 1 = (x - 3)^2

Expand the right hand side, to obtain

x - 1 = x^2 - 6x + 9

Now, move everything over to the right hand side:

0 = x^2 - 7x + 10

Factor,

0 = (x - 5)(x - 2)

This implies that x = {2, 5}. BUT WAIT! You can't assume these values will work. You have to TEST these values into the original equation, which is sqrt(x - 1) = x - 3

Test x = 2: Then, the left hand side (LHS) is equal to

LHS = sqrt(2 - 1) = sqrt(1) = 1
RHS = 2 - 3 = -1.
As you can see, LHS is NOT equal to RHS. Reject the solution
x = 2.

Test x = 5: Then
LHS = sqrt(5 - 1) = sqrt(4) = 2
RHS = 5 - 3 = 2
LHS = RHS, making this a valid solution.

Therefore, our only solution is
x = 5.

Answer 2

As you know sqrt(-1) = i
a) There seems to be something missing in problem #1, since
iSqrt(2) +18 is a complex nubeer and 4 is a real number.

b) x-3 = i implies that x= 3 + i (just add 3 to both sides)

Answer 3

The square root of a negative number is imaginary, but there is no variable to solve for in the first question.

sqrt(x-1)=x-3
Square each side.
x-1=(x-3)^2
x-1=x^2-6x+9
0=x^2-7x+10
0=(x-5)(x-2)
x=2,5

I hope that this helps.

Answer 4

I CN'T UNDERSTAND FIRST ONE.


FOR SECOND.

sqrt(x)-1=x-3
x=x^2+9-6x+1
x^2-7x+8=0

now solve this